On an Erdős Inscribed Triangle Inequality

Transcription

1 Forum Geometricorum Volume 5 (005) FORUM GEOM ISSN On an Erdős Inscribed Triangle Inequality Ricardo M. Torrejón Abstract. A comparison between the area of a triangle that of an inscribed triangle is investigated. The result obtained extend a result of Aassila giving insight into an inequality of P. Erdős. 1. Introduction Consider a triangle ABC divided into four smaller non-degenerate triangles, a central one C 1 A 1 B 1 inscribed in ABC three others on the sides of this central triangle, as depicted in C B 1 A 1 A C 1 Figure 1 B A question with a long history is that of comparing the area of ABC to that of the inscribed triangle C 1 A 1 B 1. In 1956, H. Debrunnner [5] proposed the inequality area (C 1 A 1 B 1 ) min {area (AC 1 B 1 ), area (C 1 BA 1 ), area (B 1 A 1 C)} ; (1) according to John Rainwater [7], this inequality originated with P. Erdős was communicated by N. D. Kazarinoff J. R. Isbell. However, Rainwater was more precise in stating that C 1 A 1 B 1 cannot have the smallest area of the four unless all four are equal with A 1, B 1, C 1 the midpoints of the sides BC, CA, AB. A proof of (1) first appeared in A. Bager [] later in A. Bager [3] P. H. Diana [6]. Diana s proof is particularly noteworthy; in addition to proving Erdős inequality, it also shows that the stronger form of (1) holds area (C 1 A 1 B 1 ) area (AC 1 B 1 ) area (C 1 BA 1 ) () where, without loss of generality, it is assumed that 0 < area (AC 1 B 1 ) area (C 1 BA 1 ) area (B 1 A 1 C). Publication Date: September 8, 005. Communicating Editor: Paul Yiu.

2 138 R. M. Torrejón The purpose of this paper is to show that a sharper inequality is possible when more care is placed in choosing the points A 1,B 1 C 1. In so doing we extend Aassila s inequality [1]: 4 area (A 1 B 1 C 1 ) area (ABC), which is valid when these points are chosen so as to partition the perimeter of ABC into equal length segments. Our main result is Theorem 1. Let ABC be a triangle, let A 1,B 1,C 1 be on BC, CA, AB, respectively, with none of A 1,B 1,C 1 coinciding with a vertex of ABC. If AB + BA 1 = BC + CB 1 = AC + AC 1 = α, AC + CA 1 AB + AB 1 BC + BC 1 ( ) α 1 4 area (A 1 B 1 C 1 ) area (ABC)+s 4 area (ABC) 1 where s is the semi-perimeter of ABC. When α =1we obtain Aassila s result. Corollary (Aassila [1]). Let ABC be a triangle, let A 1, B 1, C 1 be on BC, CA, AB, respectively, with none of A 1,B 1,C 1 coinciding with a vertex of ABC. If AB + BA 1 = AC + CA 1, BC + CB 1 = AB + AB 1, AC + AC 1 = BC + BC 1, 4 area (A 1 B 1 C 1 ) area (ABC).. Proof of Theorem 1 We shall make use of the following two lemmas. Lemma 3 (Curry [4]). For any triangle ABC, stard notation, 4 3 area (ABC) 9abc a + b + c. (3) Equality holds if only if a = b = c. Lemma 4. For any triangle ABC, stard notation, min{a + b + c,ab+ bc + ca} 4 3 area (ABC). (4) To prove Theorem 1, we begin by computing the area of the corner triangle AC 1 B 1 :

3 On an Erdős inscribed triangle inequality 139 C B 1 A 1 A C 1 Figure B area (AC 1 B 1 ) = 1 AC 1 AB 1 sin A = 1 AC 1 AB 1 area (ABC) AB AC = AC 1 AB AB 1 area (ABC). AC For the semi-perimeter s of ABC we have where c = AB. Also, s = AB + BC + AC = (AB + AB 1 )+(BC + CB 1 ) = ()(c + AB 1 ), AB 1 = s c s = AB + BC + AC = (AC + AC 1 )+(BC + BC 1 ) ( = 1+ 1 ) (AC + AC 1 ) α = α (b + AC 1), AC 1 = α s b

4 140 R. M. Torrejón with b = AC. Hence area (AC 1 B 1 )= 1 bc ( )( ) α s b s c area (ABC). (5) Similar computations yield area (C 1 BA 1 )= 1 ( )( ) α ca s c s a area (ABC), (6) area (B 1 A 1 C)= 1 ab ( )( ) α s a s b area (ABC). (7) From these formulae, area (A 1 B 1 C 1 ) = area (ABC) area (AC 1 B 1 ) area (C 1 BA 1 ) area (B 1 A 1 C) [ = 1 1 ( )( ) α bc s b s c 1 ( )( ) α ca s c s a 1 ( )( )] α ab s a s b area (ABC) = 1 [( )( )( ) abc s a s b s c ( )( )( )] α α α + s a s b s c area (ABC). But ( )( )( ) s a s b s c ( )( )( ) α α α + s a s b s c ( ) α 1 =(s a)(s b)(s c)+ s 3 = ( ) α 1 s [area (ABC)] + s 3. Hence ( ) abc s area (A 1 B 1 C 1 )=[area (ABC)] 3 α 1 +s 4 area (ABC). (8)

5 On an Erdős inscribed triangle inequality 141 From (3) (4) abc s 3 9 (a + b + c) area (ABC) 3 9 [a + b + c +(ab + bc + ca)] area (ABC) area (ABC) 4 area (ABC). Finally, from (8) 4 area (ABC) area (A 1 B 1 C 1 ) abc s area (A 1 B 1 C 1 ) ( ) [area (ABC)] 3 α 1 + s 4 area (ABC) a division by area (ABC) produces 4 area (A 1 B 1 C 1 ) area (ABC)+s 4 completing the proof of the theorem. ( ) α 1 [area (ABC)] 1 References [1] M. Assila, Problem 1717, Math. Mag., 78 (005) 158. [] A. Bager, Elem. Math., 1 (1957) 47. [3] A. Bager, Solution to Problem 4908, Amer. Math. Monthly, 68 (1961) [4] T. R. Curry L. Bankoff, Problem E 1861, Amer. Math. Monthly, 73 (1966) 199; solution 74 (1967) [5] H. Debrunner, Problem 60, Elem. Math., 11 (1956) 0. [6] P. H. Diana, Solution to Problem 4908, Amer. Math. Monthly, 68 (1961) 386. [7] J. Rainwater, A. Bager P. H. Dianada, Problem 4908, Amer. Math. Monthly, 67 (1960) 479. Ricardo M. Torrejón: Department of Mathematics, Texas State University San Marcos, San Marcos, Texas 78666, USA address: rt04@txstate.edu